If the sequence $\{x_n\}$ satisfies the property that $0 \leq x_{m+n} \leq x_n + x_m$ for all $n$, $m \in \mathbb{N}$ , show that the limit of the sequence $\left\{\frac{x_n}{n}\right\}_n$ exists. Provide an example of such a sequence.
I can show that it is bounded, but I don't know how to show that it is monotone increasing (I think it is monotone increasing, not sure).
Best Answer
Let's fix the number $p$ and write $n = pm + k$, where $0 \leq k < p$, then we have
\begin{align} \frac{x_{pm+k}}{pm+k} &\leq \frac{x_{pm}}{pm+k} + \frac{x_k}{pm+k} \\ & \leq \frac{mx_p}{pm} + \frac{px_1}{pm} \end{align}
when we send $m$ to infinity, we get $\limsup\frac{x_n}{n} \leq \frac{x_p}{p}$, since here $k $ can only vary finitely from 0 to $p-1$
Since $p$ is arbitrary, we can say that $\exists N_1$, such that for all $n \geq N_1$, $\frac{x_n}{n} \leq \frac{x_1}{1}$.
and $\exists N_2$, such that for all $n \geq N_2$, $\frac{x_n}{n} \leq \frac{x_{N_1}}{N_1}$
continue in this way, the sequence $\frac{x_{N_k}}{N_k}, k = 1,2,\cdots$ is decreasing and bounded for below by 0, so their limit exsits.
And the conclusion follows by $\limsup\frac{x_n}{n} \leq \frac{x_p}{p}$ for any $p$.
Or as Zarrax says in the comment, we can directly conclude by noting $\limsup\frac{x_n}{n} \leq \liminf\frac{x_p}{p}$
One example is to take $a_n = \sqrt{n}$